# Chapter – 2: Relations and Function

# Exercise 2.1

**1. If **

**, find the values of x and y.**

**Solution: –**

Given,

As the ordered pairs are equal, the corresponding elements should also be equal.

Thus,

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M and adding]

x = 2 and 3y = 3

Therefore,

x = 2 and y = 1

**2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?**

**Solution: –**

Given, set A has 3 elements and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Therefore, the number of elements in (A × B) will be 9.

**3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.**

**Solution: –**

Given, G = {7, 8} and H = {5, 4, 2}

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P × Q = {(*p*, *q*): *p *∈ P, *q* ∈ Q}

So,

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

**4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.**

**(i) If P = { m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.**

**(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs ( x, y) such that x **

**∈**

**A and**

*y*

**∈**

**B.**

**(iii) If A = {1, 2}, B = {3, 4}, then A × (B ****∩**** Φ****) =**** Φ****.**

**Solution: –**

(i) The statement is False. The correct statement is:

If P = {*m*, *n*} and Q = {*n*, *m*}, then

P × Q = {(*m*, *m*), (*m*, *n*), (*n,* *m*), (*n*, *n*)}

(ii) True

(iii) True

**5. If A = {–1, 1}, find A × A × A.**

**Solution: –**

The A × A × A for a non-empty set A is given by

A × A × A = {(*a*, *b*, *c*): *a*, *b*, *c *∈ A}

Here, It is given A = {–1, 1}

So,

A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

**6. If A × B = {( a, x), (a, y), (b, x), (b, y)}. Find A and B.**

**Solution: –**

Given,

A × B = {(*a*, *x*), (*a,* *y*), (*b*, *x*), (*b*, *y*)}

We know that the Cartesian product of two non-empty sets P and Q is given by:

P × Q = {(*p*, *q*): *p* ∈ P, *q* ∈ Q}

Hence, A is the set of all first elements and B is the set of all second elements.

Therefore, A = {*a*, *b*} and B = {*x*, *y*}

**7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that**

**(i) A × (B ****∩**** ****C) = (A ****×**** B)**** ****∩**** ****(A ****×**** C)**

**(ii) A × C is a subset of B × D**

**Solution: –**

Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

Thus,

L.H.S. = A × (B ∩ C) = A × Φ = Φ

Next,

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

Thus,

R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S

– Hence verified

(ii) To verify: A × C is a subset of B × D

First,

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

And,

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

– Hence verified

**8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.**

**Solution: –**

Given,

A = {1, 2} and B = {3, 4}

So,

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B is *n*(A × B) = 4

We know that,

If C is a set with *n*(C) = *m*, then *n*[P(C)] = 2* ^{m}*.

Thus, the set A × B has 2^{4} = 16 subsets.

And, these subsets are as below:

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

**9. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.**

**Solution: –**

Given,

*n*(A) = 3 and *n*(B) = 2; and (*x*, 1), (*y*, 2), (*z*, 1) are in A × B.

We know that,

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

So, clearly *x*, *y*, and *z* are the elements of A; and

1 and 2 are the elements of B.

As *n*(A) = 3 and *n*(B) = 2, it is clear that set A = {*x*, *y*, *z*} and set B = {1, 2}.

**10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.**

**Solution: –**

We know that,

If *n*(A) = *p *and *n*(B) = *q, *then *n*(A × B) = *pq*.

Also, *n*(A × A) = *n*(A) × *n*(A)

Given,

*n*(A × A) = 9

So, *n*(A) × *n*(A) = 9

Thus, *n*(A) = 3

Also given that, the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(*a, a*): *a* ∈ A}.

Thus, –1, 0, and 1 has to be the elements of A.

As *n*(A) = 3, clearly A = {–1, 0, 1}.

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

# Exercise 2.2

**1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {( x, y): 3x – y = 0, where x, y **

**∈**

**A}. Write down its domain, codomain and range.**

**Solution: –**

The relation R from A to A is given as:

R = {(*x*, *y*): 3*x* – *y* = 0, where *x*, *y* ∈ A}

= {(*x*, *y*): 3*x* = *y*, where *x*, *y* ∈ A}

So,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Now,

The domain of R is the set of all first elements of the ordered pairs in the relation.

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

Hence, Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

**2. Define a relation R on the set N of natural numbers by R = {( x, y): y = x + 5, x is a natural number less than 4; x, y **

**∈**

**N}. Depict this relationship using roster form. Write down the domain and the range.**

**Solution: –**

**The relation R is given by:**

R = {(*x*, *y*): *y* = *x* + 5, *x* is a natural number less than 4, *x*, *y* ∈ **N**}

The natural numbers less than 4 are 1, 2, and 3.

So,

R = {(1, 6), (2, 7), (3, 8)}

Now,

The domain of R is the set of all first elements of the ordered pairs in the relation.

Hence, Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {6, 7, 8}

**3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {( x, y): the difference between x and y is odd; x **

**∈**

**A,**

*y***∈**

**B}. Write R in roster form.**

**Solution: –**

Given,

A = {1, 2, 3, 5} and B = {4, 6, 9}

The relation from A to B is given as:

R = {(*x*, *y*): the difference between *x* and *y* is odd; *x* ∈ A, *y *∈ B}

Thus,

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

**4. The figure shows a relationship between the sets P and Q. write this relation**

**(i) in set-builder form (ii) in roster form.**

**What is its domain and range?**

**Solution: –**

From the given figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

(i) R = {(*x, y*): *y = x* – 2; *x* ∈ P} or R = {(*x, y*): *y = x* – 2 for *x* = 5, 6, 7}

Roster form

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

**5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by**

**{( a, b): a, b **

**∈**

**A,**

*b***is exactly divisible by**

*a*}.**(i) Write R in roster form**

**(ii) Find the domain of R**

**(iii) Find the range of R.**

**Solution: –**

Given,

A = {1, 2, 3, 4, 6} and relation R = {(*a*, *b*): *a*, *b* ∈ A, *b* is exactly divisible by *a*}

Hence,

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

**6. Determine the domain and range of the relation R defined by R = {( x, x + 5): x **

**∈**

**{0, 1, 2, 3, 4, 5}}.**

**Solution: –**

Given,

Relation R = {(*x*, *x* + 5): *x* ∈ {0, 1, 2, 3, 4, 5}}

Thus,

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

So,

Domain of R = {0, 1, 2, 3, 4, 5} and,

Range of R = {5, 6, 7, 8, 9, 10}

**7. Write the relation R = {( x, x**

^{3}**):**

*x*is a prime number less than 10} in roster form.**Solution: –**

Given,

Relation R = {(*x*, *x*^{3}): *x *is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

Therefore,

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

**8. Let A = { x, y, z} and B = {1, 2}. Find the number of relations from A to B.**

**Solution: –**

Given, A = {*x*, *y*, z} and B = {1, 2}.

Now,

A × B = {(*x*, 1), (*x*, 2), (*y*, 1), (*y*, 2), (*z*, 1), (*z*, 2)}

As *n*(A × B) = 6, the number of subsets of A × B will be 2^{6}.

Thus, the number of relations from A to B is 2^{6}.

**9. Let R be the relation on Z defined by R = {( a, b): a, b **

**∈**

**Z,**

*a***–**

*b*is an integer}. Find the domain and range of R.**Solution: –**

Given,

Relation R = {(*a*, *b*): *a*, *b* ∈ Z, *a *– *b* is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R = Z and Range of R = Z

# Exercise 2.3

**1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.**

**(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}**

**(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}**

**(iii) {(1, 3), (1, 5), (2, 5)}**

**Solution: –**

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

It’s seen that the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation cannot be called as a function.

**2. Find the domain and range of the following real function:**

**(i) f(x) = –|x| (ii) f(x) = √(9 – x**

^{2}**)**

**Solution: –**

(i) Given,

*f*(*x*) = –|*x*|, *x* ∈ R

We know that,

As *f*(*x*) is defined for *x* ∈ R, the domain of *f* is R.

It is also seen that the range of *f*(*x*) = –|*x*| is all real numbers except positive real numbers.

Therefore, the range of *f* is given by (–∞, 0].

(ii) f(x) = √(9 – x^{2})

As √(9 – x^{2}) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x^{2} ≥ 0.

So, the domain of *f*(*x*) is {*x*: –3 ≤ *x* ≤ 3} or [–3, 3].

Now,

For any value of *x* in the range [–3, 3], the value of *f*(*x*) will lie between 0 and 3.

Therefore, the range of *f*(*x*) is {*x*: 0 ≤ *x* ≤ 3} or [0, 3].

**3. A function f is defined by f(x) = 2x – 5. Write down the values of**

**(i) f(0), (ii) f(7), (iii) f(–3)**

**Solution: –**

Given,

Function, *f*(*x*) = 2*x* – 5.

Therefore,

(i) *f*(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) *f*(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) *f*(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

**4. The function ‘ t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.**

**Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212**

**Solution: –**

**5. Find the range of each of the following functions.**

**(i) f(x) = 2 – 3x, x **

**∈**

**R,**

*x***> 0.**

**(ii) f(x) = x**

^{2}**+ 2,**

*x*is a real number.**(iii) f(x) = x, x is a real number.**

**Solution: –**

(i) Given,

f(x) = 2 – 3*x*, *x* ∈ R, *x* > 0.

We have,

x > 0

So,

3x > 0

-3x < 0 [Multiplying by -1 both the sides, the inequality sign changes]

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

(ii) Given,

*f*(*x*) = *x*^{2} + 2, *x* is a real number

We know that,

*x*^{2} ≥ 0

So,

*x*^{2} + 2 ≥ 2 [Adding 2 both the sides]

Therefore, the value of *x*^{2} + 2 is always greater or equal to 2 for x is a real number.

Hence, Range = [2, ∞)

(iii) Given,

*f*(*x*) = *x, x* is a real number

Clearly, the range of *f* is the set of all real numbers.

Thus,

Range of *f* = R

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